Mathematics STRAIGHT LINE

PAIR OF STRAIGHT LINES

A PAIR OF STRAIGHT LINES THROUGH ORIGIN :

`(i)` A homogeneous equation of degree two of the type `ax^2 + 2hxy+ by^2 = 0` always represents a pair of
straight lines passing through the origin & if:

(a) `h^2 > ab =>` lines are real & distinct.

(b) `h^2 = ab =>` lines are coincident.

(c) `h^2 < ab =>` lines are imaginary with real point of intersection i.e. `(0, 0).`

Note: A homogeneous equation of degree n represents n straight lines through the origin.

`(ii)` lf `y = m_1 x &y = m_2x` be the two equations represented by `ax^2+ 2hxy+ by^2 = 0,` then; `m_1 + m_2 = - (2h)/b`
& `m_1m_2 = a/b.`

Explanation :

`b(y - m_1x) (y - m_2x) = ax^2+ 2hxy + by^2`

Comparing co-efficients of `x^2` and `xy` on both sides

`- b(m_1 + m_2) = 2h => m_1 + m_2 = (-2h)/b`

`b m_1 m_2 = a => m_1m_2 = a/b`

ANGLE BETWEEN STRAIGHT LINES REPRESENTED BY THE EQUATION `ax^2 + 2hxy + by^2 = 0 :`

Let the equation `ax^2+ 2hxy+ by^2 = 0` represents two lines which are `y - m_1x = 0` and `y - m_2x = 0`
`:.` Angle between these lines is

`tan theta = |(m_1-m_2)/(1+m_1m_2)| = |(sqrt((m_1+m_2)^2-4m_1m_2))/(1+m_1m_2)| = |(sqrt(((-2h)/b)^2-4a/b))/(1+a/b)| = |(2sqrt(h^2-ab))/(a+b)|`

`:.` acute angle `theta` between the pair of straight lines represented by the equation,

`ax^2 + 2hxy + by^2 = theta` is `tan theta = |(2sqrt(h^2-ab))/(a+b)|`

The condition that these lines are :

`(a)` At right angles to each other if `a + b = 0`. i.e. co- efficient of
`x^2 +` co- efficient of `y^2 = 0.`

`(b)` Coincident if `h^2 = ab.`

`(c)` Equally inclined to the axis of `x` if `h = 0`. i.e. coeff. of `xy = 0.`

BISECTORS OF ANGLE BETWEEN THE LINES REPRESENTED BY `ax^2 + 2hxy + by^2 = 0` :

Let the given equation represent the straight lines
`y - m_1x = 0` ..........(i)

`y - m_2x = 0` ........(ii)

where `m_1 + m_2 = -2h//b` and `m_1 m_2 = a/b` ........(iii)

The equation to the bisectors of the angles between the straight lines in equation (i) and (ii) are

`(y-m_1x)/(sqrt(1+m_1^2)) = pm (y-m_1x)/(sqrt(1+m_2^2))`

Therefore, the combined equation of the bisectors is

`{(y-m_1x)/(sqrt(1+m_1^2))-(y-m_1x)/(sqrt(1+m_2^2))}{(y-m_1x)/(sqrt(1+m_1^2))+(y-m_1x)/(sqrt(1+m_2^2))} = 0`

or `(y-m_1x)^2/(1+m_1^2) - (y-m_1x)^2/(1+m_2^2) = 0`

hence, by equation (iii), we get

`(-2h)/b(x^2-y^2)+2(a/b-1)xy=0`

or `(x^2-y^2)/(a-b)=(xy)/h`

Important point

The product of the perpendiculars, dropped from `(x_1, y_1)` to the pair of lines represented by the

equation, `ax^2+ 2hxy + by^2 = 0` is `(ax_1^2+2hx_1y_1+by_1^2)/sqrt((a-b)^2+4h^2)`

HOMOGENISATION

The joint equation of a pair of straight lines joining origin to the points of intersection of the line given by

`lx +my +n =0................(i)`

the 2nd degree curve : `ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0..........(ii)`

is `ax^2 +2hxy +by^2 +2gx ((lx +my)/(-n)) + 2fy((lx+my)/(-n)) +c((lx+my)/(-n))^2............(iii)`

`(iii)` is obtained by homogenizing `(ii)` with the help of `(i),` by writing `(i)` in the form : `((lx+my)/(-n)) =1`